Response:
a. 55 cars
b. 25 cars
Detailed explanation:
Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).
We find that:
N(ss) = 30
N(ac) = 30
N(sr) = 40
N(ss and ac and sr) = 15
N(at least two) = 30
a.
To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:
N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)
N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15
Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55
b.
For those cars that have exactly one feature, we have:
N(only one) = N(at least one) - N(at least two)
N(only one) = 55 - 30 = 25
1. 200% =2
5000 multiplied by 2 equals 10000
2.50% = 0.5
10000 multiplied by 0.5 equals 5000
Answer:
7 years
Step-by-step explanation:
Since a graph isn’t provided, I will indicate how many years are involved, and you should be able to deduce it from this. First, let’s [use a calculator] to find 2% of 250, which equals 5. Adding 5 to the total only slightly impacts 2% of the original amount, changing it by 0.1 or 5/50. We can continue adding 5 until we exceed 282, which occurs after 7 increments, calculated as (285 minus 250) divided by 5. Thus, the answer is indeed 7 years. A quick calculation will confirm this.
Step-by-step explanation:
a) 7!
In absence of any restrictions, the answer is 7! as it represents the permutations of all animals.
b) 4! x 3!
Considering there are 6 cats and 5 dogs, the first and last slots must be occupied by cats to ensure alternate arrangements. The only options available then are based on the arrangement of the cats among themselves and the dogs among themselves, yielding 4! permutations for the cats and 3! for the dogs, thus leading to a total of 4! x 3! arrangements.
c) 3! x 5!
Here, the arrangement of the dogs among themselves can occur in 3! ways. Considering the dogs as a singular “object,” we can arrange this unit with the 4 cats, providing 5! total arrangements possible, leading to 3! · 5! arrangement possibilities.
d) 2 x 4! x 3!
In this scenario, both cats and dogs must be grouped together, allowing positions where all cats come before the dogs or vice versa. As there are two configurations, the resultant count is 2 multiplied by both arrangements, resulting in 2 x 4! x 3!
